If the set R = {(a, b) : a + 5b = 42, a, b in | vedclass

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(B) Given $a + 5b = 42$ where $a, b \in N$.$a = 42 - 5b$.For $b = 1, a = 37$.For $b = 2, a = 32$.For $b = 3, a = 27$.For $b = 4, a = 22$.For $b = 5, a

Vedclass · Accepted Answer

$12$

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(B) Given $a + 5b = 42$ where $a, b \in N$.$a = 42 - 5b$.For $b = 1, a = 37$.For $b = 2, a = 32$.For $b = 3, a = 27$.For $b = 4, a = 22$.For $b = 5, a = 17$.For $b = 6, a = 12$.For $b = 7, a = 7$.For $b = 8, a = 2$.Thus,the set $R$ has $8$ elements,so $m = 8$.We need to calculate $\sum_{n=1}^8 (1 - i^{n!}) = x + iy$.For $n \geq 4$,$n!$ is a multiple of $4$,so $i^{n!} = (i^4)^k = 1^k = 1$.Thus,$\sum_{n=1}^8 (1 - i^{n!}) = (1 - i^{1!}) + (1 - i^{2!}) + (1 - i^{3!}) + 5(1 - 1)$.$= (1 - i) + (1 - (-1)) + (1 - (-1)) + 0$.$= 1 - i + 2 + 2 = 5 - i$.Comparing with $x + iy$,we get $x = 5$ and $y = -1$.Therefore,$m + x + y = 8 + 5 - 1 = 12$.

If the set $R = \{(a, b) : a + 5b = 42, a, b \in N\}$ has $m$ elements and $\sum_{n=1}^m (1 - i^{n!}) = x + iy$,where $i = \sqrt{-1}$,then the value of $m + x + y$ is:

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